I have been struggling to understand this weird Spark stream behaviour.
I want to write 2 files of CSV into a delta table using a Spark Streaming.
I made this example only to understand how Streams work, I dont want to use other solutions I just need to understand why is this not working.
So, I have to CSV files in /test/input:
A.csv
+---+---+
| id| x|
+---+---+
| 1| A|
| 2| B|
| 3| C|
+---+---+
B.csv
+---+---+
| id| x|
+---+---+
| 4| D|
| 5| E|
+---+---+
I read the directory (so the union of the two dataframes above) as a stream:
schema = StructType([StructField("id",IntegerType(),True), StructField("x",StringType(),True)])
df = spark.readStream.format("csv").schema(schema).option("ignoreChanges", "true").option("delimiter", ";").option("header", True).load("/test/input")
I then wanted to write this stream using the following code:
def processDf(df, epoch_id):
Ids=[x.id for x in df.select("id").distinct().collect()]
for i in Ids:
temp_df=df.filter((df.id==i))
temp_df.write.format("delta").option("mergeSchema", "true").partitionBy("id").option("replaceWhere", "id=="+str(i)).mode("append").save("/test/res")
df.writeStream.format("delta").foreachBatch(processDf).queryName("x").option("checkpointLocation", "/test/check").trigger(once=True).start()
No errors are shown. The code executes successfully.
When I go to check my files in /test/res I find all data:
But when I check delta data, I notice that only the first line is present:
df= (spark.read.format("delta").option("sep", ";").option("header", "true").load("/test/res")).cache()
+---+---+
| id| x|
+---+---+
| 1| A|
+---+---+
Why isnt it inserting all lines ? Is it the replaceWhere option ?
replaceWhere is supposed to delete only the partitions that are already in the table and got updated in source data.
What am I doing wrong please.
EDIT:
Same behaviour is noticed even if I read only one CSV in input. Code still writes only one line in output instead of all lines.
This was actually a syntaxical error, I changed the loop block with the following and it worked:
for i in ids:
i=str(i)
tmp = df.filter(df.id == i)
tmp.write.format("delta").option("mergeSchema", "true").partitionBy(PartitionKey).option("replaceWhere", "id == '$i'".format(i=i)).save("/res/")
Related
I would like to get the first and last row of each partition in spark (I'm using pyspark). How do I go about this?
In my code I repartition my dataset based on a key column using:
mydf.repartition(keyColumn).sortWithinPartitions(sortKey)
Is there a way to get the first row and last row for each partition?
Thanks
I would highly advise against working with partitions directly. Spark does a lot of DAG optimisation, so when you try executing specific functionality on each partition, all your assumptions about the partitions and their distribution might be completely false.
You seem to however have a keyColumn and sortKey, so then I'd just suggest to do the following:
import pyspark
import pyspark.sql.functions as f
w_asc = pyspark.sql.Window.partitionBy(keyColumn).orderBy(f.asc(sortKey))
w_desc = pyspark.sql.Window.partitionBy(keyColumn).orderBy(f.desc(sortKey))
res_df = mydf. \
withColumn("rn_asc", f.row_number().over(w_asc)). \
withColumn("rn_desc", f.row_number().over(w_desc)). \
where("rn_asc = 1 or rn_desc = 1")
The resulting dataframe will have 2 additional columns, where rn_asc=1 indicates the first row and rn_desc=1 indicates the last row.
Scala: I think the repartition is not by come key column but it requires the integer how may partition you want to set. I made a way to select the first and last row by using the Window function of the spark.
First, this is my test data.
+---+-----+
| id|value|
+---+-----+
| 1| 1|
| 1| 2|
| 1| 3|
| 1| 4|
| 2| 1|
| 2| 2|
| 2| 3|
| 3| 1|
| 3| 3|
| 3| 5|
+---+-----+
Then, I use the Window function twice, because I cannot know the last row easily but the reverse is quite easy.
import org.apache.spark.sql.expressions.Window
val a = Window.partitionBy("id").orderBy("value")
val d = Window.partitionBy("id").orderBy(col("value").desc)
val df = spark.read.option("header", "true").csv("test.csv")
df.withColumn("marker", when(rank.over(a) === 1, "Y").otherwise("N"))
.withColumn("marker", when(rank.over(d) === 1, "Y").otherwise(col("marker")))
.filter(col("marker") === "Y")
.drop("marker").show
The final result is then,
+---+-----+
| id|value|
+---+-----+
| 3| 5|
| 3| 1|
| 1| 4|
| 1| 1|
| 2| 3|
| 2| 1|
+---+-----+
Here is another approach using mapPartitions from RDD API. We iterate over the elements of each partition until we reach the end. I would expect this iteration to be very fast since we skip all the elements of the partition except the two edges. Here is the code:
df = spark.createDataFrame([
["Tom", "a"],
["Dick", "b"],
["Harry", "c"],
["Elvis", "d"],
["Elton", "e"],
["Sandra", "f"]
], ["name", "toy"])
def get_first_last(it):
first = last = next(it)
for last in it:
pass
# Attention: if first equals last by reference return only one!
if first is last:
return [first]
return [first, last]
# coalesce here is just for demonstration
first_last_rdd = df.coalesce(2).rdd.mapPartitions(get_first_last)
spark.createDataFrame(first_last_rdd, ["name", "toy"]).show()
# +------+---+
# | name|toy|
# +------+---+
# | Tom| a|
# | Harry| c|
# | Elvis| d|
# |Sandra| f|
# +------+---+
PS: Odd positions will contain the first partition element and the even ones the last item. Also note that the number of results will be (numPartitions * 2) - numPartitionsWithOneItem which I expect to be relatively small therefore you shouldn't bother about the cost of the new createDataFrame statement.
My parquet file is derived from CSV in which so some of the cells are escaped. Eg: This is a value
"a , ""Hello"" c"
I want this to be read by parquet as
a , "Hello" c
I am trying to escape quotes from the parquet file while reading.
If I were reading a CSV, I can do it in the following way
df = spark.read.option('quote', '"').
option('escape', '"').csv("./temp.csv")
However, we don't have a similar thing for parquet files. I've read parquet both using options and without options
>>> dfP = spark.read.parquet("./temp.parquet")
>>> dfP.show()
+---+---+---+----------------+---+
|_c0|_c1|_c2| _c3|_c4|
+---+---+---+----------------+---+
| A| B| C| D| E|
| 1| 2|3,4|"a, ""HEllo"" c"| 5|
+---+---+---+----------------+---+
>>> dfP = spark.read.option('quote', '"').
option('escape', '"').parquet("./temp.parquet")
>>> dfP.show()
+---+---+---+----------------+---+
|_c0|_c1|_c2| _c3|_c4|
+---+---+---+----------------+---+
| A| B| C| D| E|
| 1| 2|3,4|"a, ""HEllo"" c"| 5|
+---+---+---+----------------+---+
I want the D column to be read as 'a, "Hello" c'.
Is there any way to make it work?
The input parquet is converted from a CSV file which is
A,B,C,D,E
1,2,"3,4","a, ""HEllo"" c",5
EDIT: Parquet is already generated. I cannot change the way parquet is being generated. I've to use the parquet I got and try to escape.
As far as I know there is only one option for parquet files. And it is for compression. Other options like 'quote', 'delimiter', 'escape' are for csv files. So they don't work for parquet files.
I tried to simulate your case and I think that best solution for this case is using functions. Firstly I created a csv file and put it to HDFS
[ali#aliyesilli ~]$ hadoop fs -cat /test/exCsv/test.csv
A,B,C,D,E
1,2,"3,4","a, ""HEllo"" c",5
Then I read it as a csv file and save it again as a parquet file
>>> df = spark.read.csv('hdfs://localhost:8020/test/exCsv',header=True)
>>> df.show()
+---+---+---+----------------+---+
| A| B| C| D| E|
+---+---+---+----------------+---+
| 1| 2|3,4|"a, ""HEllo"" c"| 5|
+---+---+---+----------------+---+
>>> df.write.parquet('hdfs://localhost:8020/test/exPar')
When I tried to read the parguet file, D column is including double quotation as you mentioned
>>> spark.read.parquet('hdfs://localhost:8020/test/exPar').show()
+---+---+---+----------------+---+
| A| B| C| D| E|
+---+---+---+----------------+---+
| 1| 2|3,4|"a, ""HEllo"" c"| 5|
+---+---+---+----------------+---+
Then I defined a function which is called strip and use it with regexp_replace function to create string something you want to see
>>> import pyspark.sql.functions as func
>>> strip=func.udf(lambda x: x.strip('"'))
>>>
>>> spark.read.parquet('hdfs://localhost:8020/test/exPar').withColumn('D', func.regexp_replace(strip('D'), '""', '"')).show()
+---+---+---+------------+---+
| A| B| C| D| E|
+---+---+---+------------+---+
| 1| 2|3,4|a, "HEllo" c| 5|
+---+---+---+------------+---+
Maybe there are another different solutions but in this case I think you should use an udf or sql functions
This question already has answers here:
How to avoid duplicate columns after join?
(10 answers)
Closed 4 years ago.
I want to use join with 3 dataframe, but there are some columns we don't need or have some duplicate name with other dataframes, so I want to drop some columns like below:
result_df = (aa_df.join(bb_df, 'id', 'left')
.join(cc_df, 'id', 'left')
.withColumnRenamed(bb_df.status, 'user_status'))
Please note that status column is in two dataframes, i.e. aa_df and bb_df.
The above doesn't work. I also tried to use withColumn, but the new column is created, and the old column is still existed.
If you are trying to rename the status column of bb_df dataframe then you can do so while joining as
result_df = aa_df.join(bb_df.withColumnRenamed('status', 'user_status'),'id', 'left').join(cc_df, 'id', 'left')
I want to use join with 3 dataframe, but there are some columns we don't need or have some duplicate name with other dataframes
That's a fine use case for aliasing a Dataset using alias or as operators.
alias(alias: String): Dataset[T] or alias(alias: Symbol): Dataset[T]
Returns a new Dataset with an alias set. Same as as.
as(alias: String): Dataset[T] or as(alias: Symbol): Dataset[T]
Returns a new Dataset with an alias set.
(And honestly I did only now see the Symbol-based variants.)
NOTE There are two as operators, as for aliasing and as for type mapping. Consult the Dataset API.
After you've aliases a Dataset, you can reference columns using [alias].[columnName] format. This is particularly handy with joins and star column dereferencing using *.
val ds1 = spark.range(5)
scala> ds1.as('one).select($"one.*").show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
+---+
val ds2 = spark.range(10)
// Using joins with aliased datasets
// where clause is in a longer form to demo how ot reference columns by alias
scala> ds1.as('one).join(ds2.as('two)).where($"one.id" === $"two.id").show
+---+---+
| id| id|
+---+---+
| 0| 0|
| 1| 1|
| 2| 2|
| 3| 3|
| 4| 4|
+---+---+
so I want to drop some columns like below
My general recommendation is not to drop columns, but select what you want to include in the result. That makes life more predictable as you know what you get (not what you don't). I was told that our brains work by positives which could also make a point for select.
So, as you asked and I showed in the above example, the result has two columns of the same name id. The question is how to have only one.
There are at least two answers with using the variant of join operator with the join columns or condition included (as you did show in your question), but that would not answer your real question about "dropping unwanted columns", would it?
Given I prefer select (over drop), I'd do the following to have a single id column:
val q = ds1.as('one)
.join(ds2.as('two))
.where($"one.id" === $"two.id")
.select("one.*") // <-- select columns from "one" dataset
scala> q.show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
+---+
Regardless of the reasons why you asked the question (which could also be answered with the points I raised above), let me answer the (burning) question how to use withColumnRenamed when there are two matching columns (after join).
Let's assume you ended up with the following query and so you've got two id columns (per join side).
val q = ds1.as('one)
.join(ds2.as('two))
.where($"one.id" === $"two.id")
scala> q.show
+---+---+
| id| id|
+---+---+
| 0| 0|
| 1| 1|
| 2| 2|
| 3| 3|
| 4| 4|
+---+---+
withColumnRenamed won't work for this use case since it does not accept aliased column names.
scala> q.withColumnRenamed("one.id", "one_id").show
+---+---+
| id| id|
+---+---+
| 0| 0|
| 1| 1|
| 2| 2|
| 3| 3|
| 4| 4|
+---+---+
You could select the columns you're interested in as follows:
scala> q.select("one.id").show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
+---+
scala> q.select("two.*").show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
+---+
Please see the docs : withColumnRenamed()
You need to pass the name of the existing column and the new name to the function. Both of these should be strings.
result_df = aa_df.join(bb_df,'id', 'left').join(cc_df, 'id', 'left').withColumnRenamed('status', 'user_status')
If you have 'status' columns in 2 dataframes, you can use them in the join as aa_df.join(bb_df, ['id','status'], 'left') assuming aa_df and bb_df have the common column. This way you will not end up having 2 'status' columns.
I have 2 dataframes with columns as shown below.
Note: Column uid is not a unique key, and there're duplicate rows with the same uid in the dataframes.
val df1 = spark.read.parquet(args(0)).drop("sv")
val df2 = spark.read.parquet(args(1))
scala> df1.orderBy("uid").show
+----+----+---+
| uid| hid| sv|
+----+----+---+
|uid1|hid2| 10|
|uid1|hid1| 10|
|uid1|hid3| 10|
|uid2|hid1| 2|
|uid3|hid2| 10|
|uid4|hid2| 3|
|uid5|hid3| 5|
+----+----+---+
scala> df2.orderBy("uid").show
+----+----+---+
| uid| pid| sv|
+----+----+---+
|uid1|pid2| 2|
|uid1|pid1| 1|
|uid2|pid1| 2|
|uid3|pid1| 3|
|uid3|pidx|999|
|uid3|pid2| 4|
|uidx|pid1| 2|
+----+----+---+
scala> df1.drop("sv")
.join(df2, "uid")
.groupBy("hid", "pid")
.agg(count("*") as "xcnt", sum("sv") as "xsum", avg("sv") as "xavg")
.orderBy("hid").show
+----+----+----+----+-----+
| hid| pid|xcnt|xsum| xavg|
+----+----+----+----+-----+
|hid1|pid1| 2| 3| 1.5|
|hid1|pid2| 1| 2| 2.0|
|hid2|pid2| 2| 6| 3.0|
|hid2|pidx| 1| 999|999.0|
|hid2|pid1| 2| 4| 2.0|
|hid3|pid1| 1| 1| 1.0|
|hid3|pid2| 1| 2| 2.0|
+----+----+----+----+-----+
In this demo case, everything looks good.
But when I apply the same operations on the production large data, the final output contains many duplicate rows (of same (hid, pid) pair).
I though the groupBy operator would be like select distinct hid, pid from ..., but obviously not.
So what's wrong with my operation? Should I repartition the dataframe by hid, pid?
Thanks!
-- Update
And if I add .drop("uid") once I join the dataframes, then some rows are missed from the final output.
scala> df1.drop("sv")
.join(df2, "uid").drop("uid")
.groupBy("hid", "pid")
.agg(count("*") as "xcnt", sum("sv") as "xsum", avg("sv") as "xavg")
.orderBy("hid").show
To be honest I think that there are problems with the data, not the code. Of course there shouldn't be any duplicates if pid and hid are truly different (I've seen some rogue Cyrillic symbols in data before).
To debug this issue you can try and see what combinations of 'uid' and sv values represent each duplicate row.
df1.drop( "sv" )
.join(df2, "uid")
.groupBy( "hid", "pid" )
.agg( collect_list( "uid" ), collect_list( "sv" ) )
.orderBy( "hid" )
.show
After that you'll have some start point to assess your data. Or, if the lists of uid (and 'sv') are the same, file a bug.
I think I might have found the root cause.
Maybe this is caused by AWS S3 consistency model.
The background is, I submitted 2 Spark jobs to create 2 tables, and submitted a third task to join the two tables (I split them in case any of them fails and I don't need to re-run them).
I put these 3 spark-submit in a shell script running in sequence, and got the result with duplicated rows.
When I re-ran the last job just now, the result seems good.
I have a Pyspark Dataframe with this structure:
+----+----+----+----+---+
|user| A/B| C| A/B| C |
+----+----+-------------+
| 1 | 0| 1| 1| 2|
| 2 | 0| 2| 4| 0|
+----+----+----+----+---+
I had originally two dataframes, but I outer joined them using user as key, so there could be also null values. I can't find the way to sum the columns with equal name in order to get a dataframe like this:
+----+----+----+
|user| A/B| C|
+----+----+----+
| 1 | 1| 3|
| 2 | 4| 2|
+----+----+----+
Also note that there could be many equal columns, so selecting literally each column is not an option. In pandas this was possible using "user" as Index and then adding both dataframes. How can I do this on Spark?
I have a work around for this
val dataFrameOneColumns=df1.columns.map(a=>if(a.equals("user")) a else a+"_1")
val updatedDF=df1.toDF(dataFrameOneColumns:_*)
Now make the Join then the out will contain the Values with different names
Then make the tuple of the list to be combined
val newlist=df1.columns.filter(_.equals("user").zip(dataFrameOneColumns.filter(_.equals("user"))
And them Combine the value of the Columns within each tuple and get the desired output !
PS: i am guessing you can write the logic for combining ! So i am not spoon feeding !